Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. $$ f(x)=4 x^{2}-12 x-3 $$

Step-by-Step Solution

Answer

Vertex: (1.5, -12), Axis of symmetry: x = 1.5, Intercepts: y = (0, -3), x = \(1.5 \pm \sqrt{3}\).

Step 1: Identify the standard form

The quadratic function is given by \[ f(x) = 4x^2 - 12x - 3 \]This is in the standard form of a quadratic equation, which is \[ f(x) = ax^2 + bx + c \]where \( a = 4 \), \( b = -12 \), and \( c = -3 \).

Step 2: Find the vertex

The vertex of a parabola can be found using the formula for the x-coordinate:\[ x = -\frac{b}{2a} \]Substituting the given values:\[ x = -\frac{-12}{2 \times 4} = \frac{12}{8} = 1.5 \]Now substitute \( x = 1.5 \) back into the function to find the y-coordinate:\[ f(1.5) = 4(1.5)^2 - 12(1.5) - 3 = 4(2.25) - 18 - 3 = 9 - 18 - 3 = -12 \]Thus, the vertex is \( (1.5, -12) \).

Step 3: Determine the axis of symmetry

The axis of symmetry for a quadratic function is the vertical line that goes through the vertex. Thus, the axis of symmetry is found using the x-coordinate of the vertex:\[ x = 1.5 \]So the axis of symmetry is \( x = 1.5 \).

Step 4: Find the y-intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the quadratic function:\[ f(0) = 4(0)^2 - 12(0) - 3 = -3 \]Thus, the y-intercept is at \( (0, -3) \).

Step 5: Locate the x-intercepts

The x-intercepts occur when \( f(x) = 0 \). Solve the equation:\[ 4x^2 - 12x - 3 = 0 \]Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), substituting \( a = 4 \), \( b = -12 \), \( c = -3 \):\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]\[ x = \frac{12 \pm \sqrt{144 + 48}}{8} \]\[ x = \frac{12 \pm \sqrt{192}}{8} \]\[ x = \frac{12 \pm 8\sqrt{3}}{8} \]\[ x = 1.5 \pm \sqrt{3} \]Thus, the x-intercepts are at \( \left(1.5 + \sqrt{3}, 0\right) \) and \( \left(1.5 - \sqrt{3}, 0\right) \).

Step 6: Sketch the graph

The graph of the function \( f(x) = 4x^2 - 12x - 3 \) is a parabola. It opens upwards because the coefficient \( a = 4 \) is positive. The vertex is at \( (1.5, -12) \) and the axis of symmetry is \( x = 1.5 \). The y-intercept is \( (0, -3) \), and the x-intercepts are \( \left(1.5 + \sqrt{3}, 0\right) \) and \( \left(1.5 - \sqrt{3}, 0\right) \). To sketch the graph, plot the vertex, axis of symmetry, and intercepts, then draw a symmetric parabola.

Key Concepts

Vertex

A vertex is a crucial point of a quadratic function. It serves as the "turning point" of the parabola, making it either the highest or lowest point on the graph.
The vertex can be found using the formula for the x-coordinate: \( x = -\frac{b}{2a} \). Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation in standard form, \( ax^2 + bx + c \).
In our example, \( a = 4 \), \( b = -12 \), and \( c = -3 \). Plugging these into the formula gives:

• \( x = -\frac{-12}{2 \cdot 4} = 1.5 \)

To find the y-coordinate, substitute \( x \) back into the quadratic equation:
\[ f(1.5) = 4(1.5)^2 - 12(1.5) - 3 = -12 \]
This calculation reveals that the vertex is at \( (1.5, -12) \). Remember, the vertex is a key indicator of the graph's direction and shape.

Axis of Symmetry

The axis of symmetry is an imaginary vertical line that divides the parabola into two symmetrical halves.
This line passes directly through the vertex, offering a "mirror image" effect about which the parabola is symmetrical.
For any quadratic equation, the axis of symmetry can easily be found by using the x-coordinate of the vertex:

• \( x = -\frac{b}{2a} \)

In the exercise above, we've already calculated this value as \( x = 1.5 \). Hence, the axis of symmetry is the line \( x = 1.5 \).
Placing this line on your graph helps visually balance the parabola, making it easier to identify and plot other key points.

Intercepts

Intercepts are the points where the parabola intersects the x-axis and y-axis.
These points convey vital information about the behavior of the graph. In our problem, there are two types of intercepts to find:

• Y-intercept: This is where the graph crosses the y-axis (when \( x = 0 \)).
  Substitute \( x = 0 \) into the quadratic function: \( f(0) = -3 \). Thus, the y-intercept is at \( (0, -3) \).


• X-intercepts: These occur when the quadratic equation is equal to zero (\( f(x) = 0 \)).
  Solving \( 4x^2 - 12x - 3 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find two solutions:
  • \( x = 1.5 + \sqrt{3} \)
  • \( x = 1.5 - \sqrt{3} \)
  These provide the x-intercepts \( \left(1.5 + \sqrt{3}, 0\right) \) and \( \left(1.5 - \sqrt{3}, 0\right) \).

Finding these intercepts enhances the accuracy of your quadratic graph by showing where the function touches or crosses the axes.